TransWikia.com

Is symmetrization $xp-px$ required for commutation $[H,x]=0$?

Physics Asked by Qbuoy on January 23, 2021

Given a Quantum Hamiltonian: $$hat{H}=ax^2+bp^2$$ It does not commute with either $x$ or $p$. Suppose we have a Hamiltonian :$$H = k hat{p}hat{x}$$ why do we need it to be: $$H = k (hat{p}hat{x} – hat{x}hat{p})$$ And why can’t we leave it in the former form? What can be the possible reason we don’t have Hamiltonians of the format :$$H = k hat{p}hat{x}$$ and what can be the eigensolutions?

Edit1: Operators commuting with $H$ does not evolve with time by the Heisenberg Equation of motion :$$ i hbarfrac{dA_H}{dt}=[A_H,H]$$ so I wished to construct Hermitian Hamiltonians such as :$$H = k (hat{p}hat{x} – hat{x}hat{p})$$ which commute with one of the conjugating variables.

To be precise, does symmetrizing like this surely result in commuting Hamiltonians ? What if we have $hat{H}(sigma_x$,$sigma_y$,$sigma_z)$ instead of {$x,p$} ? What would be the commutable symmetric Hamiltonian and how to find its eigensolutions ?

2 Answers

I'm not sure this is the right answer, but one problem that I can immediately see is that if you had a Hamiltonian $$hat{H} = k hat{p}hat{x},$$ then it wouldn't be Hermitian, since you could show that $$hat{H}^dagger = k^* hat{x}hat{p},$$ which, since $hat{x}$ and $hat{p}$ don't commute, is not equal to $hat{H}$. Thus, $hat{H} neq hat{H}^dagger$ and so its eigenvalues needn't necessarily be real, which, to my understanding, goes against the postulates of Quantum Mechanics.

Your "symmetric" guy has a slightly weaker problem, since in that case $hat{H} = - hat{H}^dagger$, but this could be fixed by using a purely imaginary $k$, for example. Of course, a nicer "symmetric" Hamiltonian would be $hat{H} = k left(hat{x}hat{p} {color{red}+} hat{p}hat{x}right)$.

Again, I'm not sure if this is the only reason, but it certainly seems like an important one!

Answered by Philip on January 23, 2021

As it has been already pointed by others:

  • The Hamiltonian $hat{H} = khat{x}hat{p}$ is not Hermitian, i.e. it does not correspond to anything measurable.
  • One can obtain Hermitian Hamiltonians by using either symmetrizing or anti-symmetrizing the operator product $hat{x}hat{p}$: $$hat{H}_+ = frac{k}{2}(hat{x}hat{p} + hat{p}hat{x}), hat{H}_- = frac{-ik}{2}(hat{x}hat{p} - hat{p}hat{x})$$ (Of course, in this particular case the second option is trivial, since the commutator is $ihbar$.)

Now my remark:
The symmetrized version of this Hamiltonian is far from exotic - it is the Hamiltonian of a particle in magnetic field: $$hat{H}=frac{1}{2m}left(hat{mathbf{p}} -frac{e}{c}mathbf{A}(mathbf{r})right)^2 = frac{1}{2m}left[hat{mathbf{p}}^2 -frac{e}{c}left(hat{mathbf{p}}mathbf{A}(mathbf{r}) + mathbf{A}(mathbf{r})hat{mathbf{p}}right) + frac{e^2}{c^2}mathbf{A}^2(mathbf{r})right]$$ While in QM books the discussion is usually limited to Landau gauge, where the vector potential and the momentum commute, in general this is not the case.

Answered by Vadim on January 23, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP